mình có mấy bài nè:
2. let x, y be real number such that 4[TEX]x^2[/TEX] + [TEX]y^2[/TEX] = 1. Find the maximum and minimum values of the expression
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]
NHỚ THANKS MÌNH NHA
Dịch:
2. Cho x, y là số thực đảm bảo điều kiện 4[TEX]x^2[/TEX] + [TEX]y^2[/TEX] = 1. Tìm giá trị lớn nhất và nhỏ nhất của biểu thức.
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX] ( * )
Giải:
Based on the condition: 4[TEX]x^2[/TEX] + [TEX]y^2[/TEX] = 1
The left side of this condition is always positive because:
4[TEX]x^2[/TEX] and [TEX]y^2[/TEX] \geq 0. Therefore, the above-mentioned condition is true when
(#) 4[TEX]x^2[/TEX] =1 and [TEX]y^2[/TEX] = 0
(##) 4[TEX]x^2[/TEX] =0 and [TEX]y^2[/TEX] = 1
From (#), with y=0 and 4[TEX]x^2[/TEX]=1, there will be two cases. (x = 0.5 and x = -0.5)
1) x=0.5 and y =0; ==> the value of the expression will be :
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]= [TEX]\frac{1}{3}[/TEX]=1/3
2)x=-0.5 and y =0; ==> the value of the expression will be :
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]= [TEX]\frac{-1}{1}[/TEX]=-1
From (##), there will be two cases. (y =1 or y =-1)
3) x=0, y = 1; ==> the value of the expression will be :
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]= [TEX]\frac{3}{3}[/TEX]=1
4) x=0, y = -1; ==> the value of the expression will be:
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]= [TEX]\frac{-3}{1}[/TEX]=-3
Conclusion:
- Comparing the results from 1), 2), 3), 4), the maximum value of the expression:
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]= 1, when x=0 and y = 1; and the minimum value of the expression:
A = [TEX]\frac{2x + 3y}{2x + y + 2}[/TEX]= -3, when x=0 and y = -1.